CHI035 power board capacitor heat

[satrapus]

I have modified the CHI035 power board so that it can be powered by a single car battery (12.8v when fully charged).
I followed the steps stated in the pdf for the mods and the board works fine, however I noticed that the sole capacitor
on the board is getting really hot.
As I need the board to be reliable is there anything I can do to remedy this?
From what little I understand about electronics the capacitor will not last very long if it's overheating.
Can I replace it with a capacitor with different specs and if so which one would be appropriate?

Many thanks

 

[nick12ab]

What is the voltage rating of the capacitor?
is it the capacitor that gets hot or the voltage regulator next to it?
Is the capacitor connected the right way round?
Did it get hot when it was powered by a lower voltage source?

 

[eggdweather]

It may be faulty. Usually capacitors start to heat when current flows due to leakage of the electrolyte. A capacitor should have near infinite resistance between the pins/internal plates so there can be no resistive current flow between the plates, however for large ac currents flowing in and out of the capacitor there would be current flowing, but the currents would need to be very high to get a heating effect.
Is your load 'complex' that is creating a lot of electrical and audible noise (whining) suggesting lots of ac components or is it a simple resistive load. What I would do it try your board with a resistive load/light bulb and see it still gets hot, either way, the most likely cause is a faulty capacitor. Your replacement will likely be limited to the space available to you.

 

[satrapus]

Bit tricky to read but I think the capacitor provided on the board is rated 15 volts.
Now that I checked, the voltage regulator next to it is much hotter than the capacitor itself.
Given their proximity of the two there may not be an issue with the capacitor after all.
I'll leave the board powered up for a couple of days and see what happens.

Not too sure about load complexity ( noob ) On the 12v output side of the board I have a low rpm dc motor and a buzzer.
On the input side a ir proximity sensor, pot and 3 way switch.

 

[nick12ab]

Bit tricky to read but I think the capacitor provided on the board is rated 15 volts...

The voltage regulator is the likely heat source, though using a capacitor with a voltage rating just above what the battery voltage will be when charging is quite dubious. I'd suggest upgrading the capacitor to a 25V one.

As for the voltage regulator, the heat is caused by high current consumption (which for a small linear regulator like that could be as little as 50mA) so what is this proximity sensor you're using? You may need to use a switching regulator to power it.

 

[satrapus]

I think part of the problem is that the capacitor and regulator are so close together.
I might replace the regulator and this time around I won't snip its 3 legs so short so that it's much higher on the board.


This is the cheap ir sensor I'm using :
http://www.ebay.co.uk/itm/Iinfrared-IR-Photoelectric-Distance-Avoidance-Sensor-Adjustable-/171741780506?pt=LH_DefaultDomain_3&hash=item27fc9b9a1a

Thanks

 

[erco]

Try a big heat sink and a fan if needed. Linear regulators just dissipate energy as heat, very wasteful. They get hotter the more voltage they drop, and the more current flowing through them.

Switching regulators are much more efficient. http://www.protostack.com/semiconductors/voltage-regulators/switching-regulator-lm7805-drop-in-replacement

 

[srnet]

I might replace the regulator and this time around I won't snip its 3 legs so short so that it's much higher on the board.

If its getting hot, it needs a heatsink.

 

[Goeytex]

What regulator did you fit on the board? The CHI035 Datasheet specifies a 78L05 in TO-92 Package . If you followed the datasheet and installed the 78L05, then there are better choices. The way this regulator dissipates heat is by air moving across the plastic package and through the legs to the copper traces below. If the Picaxe and anything else using this regulator are drawing any kind of load, the heat sinking is likely to be inadequate. The fact that it is getting hot is a good indicator. While technically an LDO (Low Dropout Regulator), by today's more current offerings it is fairly inefficient. If you insist upon using the L7805 then I suggest you use a T0-92 heat sink. These are relatively cheap and will help dissipate the heat. Avid makes a slip on heat sink for TO-92 Devices.

A better choice might be to use a more efficient LDO regulator. Something like a L4931CZ50-AP. This is a 5V LDO in a TO-92 package with all around better performance than the L7805. It has better line and load regulation, a much lower dropout voltage and can supply more than twice the current as the L7805. With a slip on heat sink it should be more than adequate as an on board regulator. I see no need to use a switcher in your application, especially if you need the Picaxe regulator on the CHI035 board.

Good Luck

 

[srnet]

With a fixed input of 12.8V, in what way is one LDO more efficient than another ?

Some better regulators do have a very low quiescent current, in the region of uA.

 

[nick12ab]

This is the cheap ir sensor I'm using :
http://www.ebay.co.uk/itm/Iinfrared-IR-Photoelectric-Distance-Avoidance-Sensor-Adjustable-/171741780506?pt=LH_DefaultDomain_3&hash=item27fc9b9a1a

According to the eBay listing, that sensor consumes 100mA, which is the current rating of the 78L05. Then the other components you're powering from 5V add to that as well so you're exceeding the current limit of the 78L05.

That could just be the peak current but even so the current regulator has already been deemed inadequate and at a very minimum needs to be replaced with a TO220 package regulator. As already said, an LDO regulator will not be any better.

 

[techElder]

Not trying to be snotty, but the very first question in troubleshooting this problem should have been, "What is your load's current consumption?"

Then understand that the regulator's power dissipation is calculated from that current and the difference between 12.8 volts (of the battery) and 5 volts of the regulator.

If this value is more than the rating of the regulator, then "Bob's your uncle!". In other words, you have exceeded the capability of your regulator and must get a different, higher wattage regulator.

 

[satrapus]

Sorry for the delay in answering, I was away travelling...

I am not powering any other components from 5V apart from the ir sensor (just have a pot and a 3 way switch on 5V).
The buzzer and dc motor are on the 12V side.
I just measured the temperature on the board components.
78L05 reading up to 59c and the capacitor goes up to 40c.
Given this should I replace the 78L05 with a higher wattage regulator?
I'm not too familiar with electronic concepts, I just stuck the 78L05 on the board as it was recommended in the pdf.

Many thanks for all your advice fellas

 

[techElder]

satrapus, you really have to learn something about "electronic concepts" if you are going to be successful with your projects.

Look up the datasheet on the 78L05 and see if you have exceeded its recommended limits. The temperature of the components in question have little bearing on the solution.

In post #12, I spoke of one way to determine the correct ratings for the components in your project.